Mixing
Problem 9 - Chapter 5 - Evans' PDE (First Edition)
$begingroup$
In the $1$st edition, this was question $5.9$. The question is:
Integrate by parts to prove:
$$int_{U} |Du|^p dx leq C left(int_{U} |u|^p dxright)^{1/2} left(int_{U} |D^2 u|^p dxright)^{1/2}$$
for $ 2 leq p < infty$ and all $u in W^{2,p}(U) cap W^{1,p}_{0}(U)$.
Hint:
$$int_{U} |Du|^p dx = sum_{i=1}^{n}int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
I am trying to prove this holds for $u in C^{infty}_{c} (U)$, and then by density conclude the theorem. However, this integration does not appear to be working.
The question says to use the integration by parts formula. Is it possible to use integration by parts in the following integral?
$$int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
My friend used it here, but he doesn't know if it is valid...
Can someone please give a hint for how I could start?
pde sobolev-spaces integral-inequality
edited Feb 1 at 0:42
Pedro
10.9k23475
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
$endgroup$
add a comment |
$begingroup$
In the $1$st edition, this was question $5.9$. The question is:
Integrate by parts to prove:
$$int_{U} |Du|^p dx leq C left(int_{U} |u|^p dxright)^{1/2} left(int_{U} |D^2 u|^p dxright)^{1/2}$$
for $ 2 leq p < infty$ and all $u in W^{2,p}(U) cap W^{1,p}_{0}(U)$.
Hint:
$$int_{U} |Du|^p dx = sum_{i=1}^{n}int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
I am trying to prove this holds for $u in C^{infty}_{c} (U)$, and then by density conclude the theorem. However, this integration does not appear to be working.
The question says to use the integration by parts formula. Is it possible to use integration by parts in the following integral?
$$int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
My friend used it here, but he doesn't know if it is valid...
Can someone please give a hint for how I could start?
pde sobolev-spaces integral-inequality
edited Feb 1 at 0:42
Pedro
10.9k23475
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
$endgroup$
$begingroup$
in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question
$endgroup$
– math student
Apr 23 '13 at 3:34
$begingroup$
In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $pne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help...
$endgroup$
– math101
Jul 10 '15 at 13:16
add a comment |
$begingroup$
In the $1$st edition, this was question $5.9$. The question is:
Integrate by parts to prove:
$$int_{U} |Du|^p dx leq C left(int_{U} |u|^p dxright)^{1/2} left(int_{U} |D^2 u|^p dxright)^{1/2}$$
for $ 2 leq p < infty$ and all $u in W^{2,p}(U) cap W^{1,p}_{0}(U)$.
Hint:
$$int_{U} |Du|^p dx = sum_{i=1}^{n}int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
I am trying to prove this holds for $u in C^{infty}_{c} (U)$, and then by density conclude the theorem. However, this integration does not appear to be working.
The question says to use the integration by parts formula. Is it possible to use integration by parts in the following integral?
$$int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
My friend used it here, but he doesn't know if it is valid...
Can someone please give a hint for how I could start?
pde sobolev-spaces integral-inequality
edited Feb 1 at 0:42
Pedro
10.9k23475
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
$endgroup$
In the $1$st edition, this was question $5.9$. The question is:
Integrate by parts to prove:
$$int_{U} |Du|^p dx leq C left(int_{U} |u|^p dxright)^{1/2} left(int_{U} |D^2 u|^p dxright)^{1/2}$$
for $ 2 leq p < infty$ and all $u in W^{2,p}(U) cap W^{1,p}_{0}(U)$.
Hint:
$$int_{U} |Du|^p dx = sum_{i=1}^{n}int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
I am trying to prove this holds for $u in C^{infty}_{c} (U)$, and then by density conclude the theorem. However, this integration does not appear to be working.
The question says to use the integration by parts formula. Is it possible to use integration by parts in the following integral?
$$int_{U} u_{x_i}u_{x_i}|Du|^{p -2} dx$$
My friend used it here, but he doesn't know if it is valid...
Can someone please give a hint for how I could start?
pde sobolev-spaces integral-inequality
pde sobolev-spaces integral-inequality
edited Feb 1 at 0:42
Pedro
10.9k23475
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
edited Feb 1 at 0:42
Pedro
10.9k23475
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
edited Feb 1 at 0:42
Pedro
10.9k23475
edited Feb 1 at 0:42
Pedro
10.9k23475
edited Feb 1 at 0:42
Pedro
10.9k23475
10.9k23475
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
asked Apr 23 '13 at 3:32
math studentmath student
2,41111018
2,41111018
$begingroup$
in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question
$endgroup$
– math student
Apr 23 '13 at 3:34
$begingroup$
In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $pne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help...
$endgroup$
– math101
Jul 10 '15 at 13:16
add a comment |
$begingroup$
in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question
$endgroup$
– math student
Apr 23 '13 at 3:34
$begingroup$
In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $pne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help...
$endgroup$
– math101
Jul 10 '15 at 13:16
$begingroup$
in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question
$endgroup$
– math student
Apr 23 '13 at 3:34
$begingroup$
in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question
$endgroup$
– math student
Apr 23 '13 at 3:34
$begingroup$
In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $pne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help...
$endgroup$
– math101
Jul 10 '15 at 13:16
$begingroup$
In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $pne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help...
$endgroup$
– math101
Jul 10 '15 at 13:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you guessed, the first step is an integration by parts:
$$int_U |Du|^p dx = int_U nabla u cdot nabla u |Du|^{p-2} dx = - int_U u nabla cdot ( nabla u |Du|^{p-2}) dx = - int_U u left( Delta u |Du|^{p-2} + (p-2) (nabla u^T D^2 u nabla u) |Du|^{p-4})right) leq C int_U u |Du|^{p-2} |D^2 u| dx $$
Now the next step is an invocation of the Holder inequality, with conjugate exponents $frac{p}{2}$ and $frac{p}{p-2}$ (notice that this step requires $p gt 2$, if $p=2$, it is unnecessary):
$$int_U u |Du|^{p-2} |D^2 u| dx leq left(int_U |u|^frac{p}{2} |D^2u|^frac{p}{2}right)^frac{2}{p} left(int_U |Du|^p right)^frac{p-2}{p} $$
So, dividing the original left hand side by the gradient term and invoking Holder on the remaining term with exponent 2, we get the desired result.
$$ int_U |Du|^p dx leq C left(int_U |u|^p dx right)^frac{1}{2} left( int_U |D^2u|^p dx right)^frac{1}{2} $$
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
$endgroup$
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
|
show 1 more comment
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1 Answer
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$begingroup$
As you guessed, the first step is an integration by parts:
$$int_U |Du|^p dx = int_U nabla u cdot nabla u |Du|^{p-2} dx = - int_U u nabla cdot ( nabla u |Du|^{p-2}) dx = - int_U u left( Delta u |Du|^{p-2} + (p-2) (nabla u^T D^2 u nabla u) |Du|^{p-4})right) leq C int_U u |Du|^{p-2} |D^2 u| dx $$
Now the next step is an invocation of the Holder inequality, with conjugate exponents $frac{p}{2}$ and $frac{p}{p-2}$ (notice that this step requires $p gt 2$, if $p=2$, it is unnecessary):
$$int_U u |Du|^{p-2} |D^2 u| dx leq left(int_U |u|^frac{p}{2} |D^2u|^frac{p}{2}right)^frac{2}{p} left(int_U |Du|^p right)^frac{p-2}{p} $$
So, dividing the original left hand side by the gradient term and invoking Holder on the remaining term with exponent 2, we get the desired result.
$$ int_U |Du|^p dx leq C left(int_U |u|^p dx right)^frac{1}{2} left( int_U |D^2u|^p dx right)^frac{1}{2} $$
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
$endgroup$
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
|
show 1 more comment
$begingroup$
As you guessed, the first step is an integration by parts:
$$int_U |Du|^p dx = int_U nabla u cdot nabla u |Du|^{p-2} dx = - int_U u nabla cdot ( nabla u |Du|^{p-2}) dx = - int_U u left( Delta u |Du|^{p-2} + (p-2) (nabla u^T D^2 u nabla u) |Du|^{p-4})right) leq C int_U u |Du|^{p-2} |D^2 u| dx $$
Now the next step is an invocation of the Holder inequality, with conjugate exponents $frac{p}{2}$ and $frac{p}{p-2}$ (notice that this step requires $p gt 2$, if $p=2$, it is unnecessary):
$$int_U u |Du|^{p-2} |D^2 u| dx leq left(int_U |u|^frac{p}{2} |D^2u|^frac{p}{2}right)^frac{2}{p} left(int_U |Du|^p right)^frac{p-2}{p} $$
So, dividing the original left hand side by the gradient term and invoking Holder on the remaining term with exponent 2, we get the desired result.
$$ int_U |Du|^p dx leq C left(int_U |u|^p dx right)^frac{1}{2} left( int_U |D^2u|^p dx right)^frac{1}{2} $$
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
$endgroup$
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
|
show 1 more comment
$begingroup$
As you guessed, the first step is an integration by parts:
$$int_U |Du|^p dx = int_U nabla u cdot nabla u |Du|^{p-2} dx = - int_U u nabla cdot ( nabla u |Du|^{p-2}) dx = - int_U u left( Delta u |Du|^{p-2} + (p-2) (nabla u^T D^2 u nabla u) |Du|^{p-4})right) leq C int_U u |Du|^{p-2} |D^2 u| dx $$
Now the next step is an invocation of the Holder inequality, with conjugate exponents $frac{p}{2}$ and $frac{p}{p-2}$ (notice that this step requires $p gt 2$, if $p=2$, it is unnecessary):
$$int_U u |Du|^{p-2} |D^2 u| dx leq left(int_U |u|^frac{p}{2} |D^2u|^frac{p}{2}right)^frac{2}{p} left(int_U |Du|^p right)^frac{p-2}{p} $$
So, dividing the original left hand side by the gradient term and invoking Holder on the remaining term with exponent 2, we get the desired result.
$$ int_U |Du|^p dx leq C left(int_U |u|^p dx right)^frac{1}{2} left( int_U |D^2u|^p dx right)^frac{1}{2} $$
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
$endgroup$
As you guessed, the first step is an integration by parts:
$$int_U |Du|^p dx = int_U nabla u cdot nabla u |Du|^{p-2} dx = - int_U u nabla cdot ( nabla u |Du|^{p-2}) dx = - int_U u left( Delta u |Du|^{p-2} + (p-2) (nabla u^T D^2 u nabla u) |Du|^{p-4})right) leq C int_U u |Du|^{p-2} |D^2 u| dx $$
Now the next step is an invocation of the Holder inequality, with conjugate exponents $frac{p}{2}$ and $frac{p}{p-2}$ (notice that this step requires $p gt 2$, if $p=2$, it is unnecessary):
$$int_U u |Du|^{p-2} |D^2 u| dx leq left(int_U |u|^frac{p}{2} |D^2u|^frac{p}{2}right)^frac{2}{p} left(int_U |Du|^p right)^frac{p-2}{p} $$
So, dividing the original left hand side by the gradient term and invoking Holder on the remaining term with exponent 2, we get the desired result.
$$ int_U |Du|^p dx leq C left(int_U |u|^p dx right)^frac{1}{2} left( int_U |D^2u|^p dx right)^frac{1}{2} $$
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
edited Apr 23 '13 at 13:22
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
answered Apr 23 '13 at 11:24
Ray YangRay Yang
1,9831016
1,9831016
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
|
show 1 more comment
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
why i can use integration by parts in the integral ( the second integral in the first line of your solution) ? thank you
$endgroup$
– math student
Apr 23 '13 at 13:17
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
oops, that was a typo, since corrected. Is it clearer now?
$endgroup$
– Ray Yang
Apr 23 '13 at 13:22
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
@RayYang: May I ask why the differential of $|Du|^{p-2}|$ is so complicated..?
$endgroup$
– Bombyx mori
Apr 24 '13 at 6:46
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
It's not that complicated. $frac{partial}{partial x_i} |Du|^{p-2} = (p-2) |Du|^{p-4} sum_j frac{partial^2 u}{partial x_i partial x_j} frac{partial u}{partial x_j}$. Which turns into the form I've written above after sorting terms.
$endgroup$
– Ray Yang
Apr 24 '13 at 10:28
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
$begingroup$
@RayYang I have difficulties in this calculation, I would appreciate if you could check my question at math.stackexchange.com/questions/1355970/…
$endgroup$
– math101
Jul 10 '15 at 3:41
|
show 1 more comment
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$begingroup$
in the case p=2 , I proved first for smooth u with compact suport , and by density I proved the question
$endgroup$
– math student
Apr 23 '13 at 3:34
$begingroup$
In case $p=2$, the solution can be found in page 6 of columbia.edu/~la2462/Problems%20from%20Evans.pdf But I have some confusion when comparing with the general case (I mean $pne 2$). In $p=2$ case, one needs to approximate by smooth function...(see bottom half of page 6), but general $p$ we do not need this step, could you please explain? @math student or anyone can help...
$endgroup$
– math101
Jul 10 '15 at 13:16